# Solid formed after revolving the area between y = x^2 and y = x+2 around y axis?

## Two functions $y = {x}^{2}$ and $y = x + 2$ Around y axis It will be (16π)/3 because there is a solid inside other or we have to considerer solid 1 plus solid 2 (5π)/6 + (16π)/3 = (37π)/6

Aug 6, 2018

graph{(y-x^2)(y-x-2)=0 [-7.37, 10.41, -1.955, 6.934]}

Ignore the ${x}^{2} \le y \le x + 2$ in Q2 as it will be within the solid created by revolving the part in Q1 about the y axis

That leaves:

$\underbrace{\pi {\int}_{0}^{4} {x}^{2} \setminus \boldsymbol{\mathrm{dy}}} {\setminus}_{\text{Volume inside Parabola")- underbrace( pi r^2 h/3)_("Volume of Cone}}$

With $y = {x}^{2} q \quad \setminus \mathrm{dy} = 2 x \mathrm{dx}$

$= 4 \pi {\int}_{0}^{2} {x}^{3} \setminus \boldsymbol{\mathrm{dx}} - {\underbrace{\pi \cdot {2}^{2} \cdot \frac{2}{3}}}_{\text{Volume of Cone}}$

$= \pi {\left[{x}^{4}\right]}_{0}^{2} - \pi \cdot {2}^{2} \cdot \frac{2}{3}$

$= \frac{40}{3} \pi$